chore(Java): add Count Set Bits (#600)

algorithms/Java/README.md

@@ -83,3 +83,6 @@
 
 ## Backtracking
 - [N Queen Problem](backtracking/nqueen.java)
+
+## Bit Manipulation
+- [Count Set Bits](bit-manipulation/count-set-bits.java)

algorithms/Java/bit-manipulation/count-set-bits.java

@@ -0,0 +1,83 @@
+/** Problem Description :
+ * You are given a number N. Find the total count of set bits for all numbers from 1 to N
+ * including 1 and N
+ */
+
+
+/**
+ * Author : Suraj Kumar
+ * Github : https://github.com/skmodi649
+ */
+
+
+/** ALGORITHM :
+ * So, we will iterate till the number of bits in the number.
+ * And we don’t have to iterate every single number in the range from 1 to n.
+ * We will perform the following operations to get the desired result.
+ * The quotient when n+1 is divided by 2 will return the number of times the 0,1 pattern has appeared at LSB.
+ * However, the quotient when n+1 is divided by 4 will return the number of times the 0,0,1,1 pattern has
+ * appeared at 2nd least significant bit and so on.
+ */
+
+
+/** TEST CASES :
+ * Test Case 1 :
+ * Input: N = 4
+ * Output: 5
+ *
+ * Test Case 2 :
+ * Input: N = 17
+ * Output: 35
+ */
+
+import java.util.*;
+import java.lang.*;
+
+
+class Solution {
+
+    //Function to return sum of count of set bits in the integers from 1 to n.
+    public static int countSetBits(int n) {
+        int count = 0 , i = 1 , val = 0;
+        while ((n + 1) / (int) Math.pow(2, i) != 0) {
+            int k = (n + 1) % (int) Math.pow(2, i);
+            val = (n + 1) / (int) Math.pow(2, i);
+            int c = (int) Math.pow(2, i - 1);
+            count = count + c * val;
+            if (k > ((int) Math.pow(2, i) / 2)) {
+                count = count + (k - ((int) Math.pow(2, i) / 2));
+            }
+            i++;
+        }
+        int temp = (int) Math.pow(2, i);
+        count += (n + 1 - temp / 2);
+        return count;
+    }
+
+    public static void main(String[] args){
+        Scanner sc = new Scanner(System.in);
+        System.out.println("Enter the number :");
+            int n = sc.nextInt();//input n
+
+            Solution obj = new Solution();
+            System.out.println("Total set bits : "+obj.countSetBits(n)); // calling countSetBits method
+            System.out.println();        // changing the line
+        }
+    }
+
+
+
+/** Explanation for N = 4
+ * For numbers from 1 to 4.
+ * For 1: 0 0 1 = 1 set bits
+ * For 2: 0 1 0 = 1 set bits
+ * For 3: 0 1 1 = 2 set bits
+ * For 4: 1 0 0 = 1 set bits
+ * Therefore, the total set bits is 5.
+  */
+
+
+/** Time Complexity : O(log N)
+* Auxiliary Space Complexity : O(1)
+* Constraints : 1 ≤ N ≤ 10^8
+*/